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>> No.11637728 [View]
File: 224 KB, 1831x756, s22.png [View same] [iqdb] [saucenao] [google]
11637728

>>11637630
Are you still in high school? It's not happening otherwise.

>> No.11584525 [View]
File: 224 KB, 1831x756, s22.png [View same] [iqdb] [saucenao] [google]
11584525

>>11568541
https://en.wikipedia.org/wiki/Category_of_small_categories
>>11569182
It should be [math]a<x<c<y<b[/math], yes. On both cases.
The proof is otherwise correct.
>>11573098
Recalling that [math]sup ~ f \leq M[/math] and [math]inf ~ f \geq - M \rightarrow M \geq -inf ~ f[/math] we get [math]sup ~ f - inf ~ f \leq M + M = 2M[/math].
>>11576357
You now have [math]\frac{5}{4}[/math] times what you had earlier, so it's a 25% increase.
>>11576521
Baby don't hurt me.
>>11576533
The equivalence part is definitely correct.
The issue is the "relation" part, since even if you exclude [math](0, 0)[/math], you still have issues with division by zero.
Unless you go and consider undefined equal to undefined or similar finicky stuff.
Was it supposed to be [math](x_1, y_1) R (x_2, y_2)[/math] whenever [math]y_1 x_2 = y_2 x_1[/math] or something similar?
>>11576539
Yeah, seems about right.
>>11582978
The computation seems overall correct, but the interval of convergence looks wonky.
Consider that [math]\frac{1}{(1-x)^2} = ( \sum _{n=0}^{\infty} x^n )^2[/math], which tells you that the radius of convergence includes that of [math]\sum _{n=0}^{\infty} x^n[/math] and that this sum dominates [math]\sum _{n=0}^{\infty} x^n[/math] term-wise for [math]x> 1[/math].
>>11583487
Seems correct.
Just sub in u in terms of x.

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